Question

If $f\left(0\right)=f\left(1\right)=0,f\text{'}\left(1\right)=2\mathrm{and}y=f\left(e^x\right)e^{f\left(x\right)}$, write the value of $\frac{dy}{dx}\mathrm{at}x=0$.

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have},f\left(0\right)=f\left(1\right)=0,f\text{'}\left(1\right)=2 \\ \mathrm{and}, \\ y=f\left(e^x\right)e^{f\left(x\right)} \end{gathered}$$

$$\begin{gathered} \Rightarrow \frac{dy}{dx}=\frac{d}{dx}\left[f\left(e^x\right)\times e^{f\left(x\right)}\right] \\ \Rightarrow \frac{dy}{dx}=f\left(e^x\right)\frac{d}{dx}{e}^{f\left(x\right)}+e^{f\left(x\right)}\frac{d}{dx}f\left(e^x\right)\left[\mathrm{Using}\mathrm{product}\mathrm{rule}\right] \\ \Rightarrow \frac{dy}{dx}=f\left(e^x\right)\times e^{f\left(x\right)}\frac{d}{dx}f\left(x\right)+e^{f\left(x\right)}\times f\text{'}\left(e^x\right)\frac{d}{dx}\left(e^x\right) \\ \Rightarrow \frac{dy}{dx}=f\left(e^x\right)\times e^{f\left(x\right)}\times f\text{'}\left(x\right)+e^{f\left(x\right)}\times f\text{'}\left(e^x\right)\times e^x \\ \mathrm{Putting}x=0,\mathrm{we}\mathrm{get}, \\ \frac{dy}{dx}=f\left(e^0\right)\times e^{f\left(0\right)}\times f\text{'}\left(0\right)+e^{f\left(0\right)}\times f\text{'}\left(e^0\right)\times e^0 \\ \Rightarrow \frac{dy}{dx}=f\left(1\right)e^{f\left(0\right)}\times f\text{'}\left(0\right)+e^{f\left(0\right)}\times f\text{'}\left(1\right)\times 1 \\ \Rightarrow \frac{dy}{dx}=0\times e^0\times f\text{'}\left(0\right)+e^0\times 2\times 1\left[\because f\left(x\right)=f\left(1\right)=0\mathrm{and}f\text{'}\left(1\right)=2\right] \\ \Rightarrow \frac{dy}{dx}=0+1\times 2\times 1 \\ \Rightarrow \frac{dy}{dx}=2 \end{gathered}$$